Problem: Rewrite the function by completing the square. $f(x)=x^{2}-8x-2$ $f(x)=(x+$
Solution: We want to complete $x^2{-8}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-8}}{2}\right)^2={16}$ to it: $x^2{-8}x+{16}=(x-4)^2$ In order to keep the expression equivalent, we add and subtract ${16}$, not forgetting the expression's constant term, $-2$ : $\begin{aligned} f(x)&=x^2-8x-2 \\\\ &=x^2-8x+{16}-2-{16} \\\\ &=(x-4)^2-2-16 \\\\ &=(x-4)^2-18 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 4)^2 - 18$ This is equivalent to $f(x)=(x+{-4})^2-18$